Logarithms: UFRGS 2017 Exam Explained

Hey guys, welcome back! Today, we're diving deep into a super interesting problem from the UFRGS 2017 exam, specifically question 5, involving logarithms. Now, I know logarithms can sometimes feel like a bit of a headache, but trust me, once you get the hang of the core concepts, they become way more manageable, and honestly, pretty cool. We're going to break down this specific problem, log₅(x²) = 2, step-by-step, so you can see exactly how to approach it. My goal here is to make sure you not only understand how to solve this but also why each step is necessary. We’ll be talking about the fundamental properties of logarithms, how to convert logarithmic equations into exponential ones, and how to deal with variables inside the logarithm. So, grab your notebooks, get comfy, and let's get this math party started!

Understanding the Logarithmic Equation: log₅(x²) = 2

Alright, let's get down to business with our problem: log₅(x²) = 2. Before we jump into solving for x, it's crucial to understand what this equation is actually telling us. In simple terms, a logarithm answers the question: "To what power must we raise the base to get a certain number?" In our case, the base is 5. The equation log₅(x²) = 2 is asking: "To what power must we raise 5 to get x²?" And the answer to that question, according to the equation, is 2. So, we're looking for a value of x such that when you square it, and then take the logarithm base 5 of that result, you end up with 2. Pretty straightforward when you break it down like that, right?

Now, let's think about the properties of logarithms that might come in handy here. One of the most powerful properties is the power rule, which states that log_b(M^p) = p * log_b(M). This rule allows us to bring an exponent down as a multiplier. In our equation, log₅(x²) = 2, we have x² inside the logarithm. Applying the power rule, we can rewrite this as 2 * log₅(x) = 2. This transformation is super useful because it isolates the logarithm of x itself, making it easier to solve for x. Remember, math is all about using the right tools and properties to simplify complex expressions. This is exactly what we're doing here. We're not changing the value of the expression; we're just changing its form to make it more approachable.

Another way to tackle this is by converting the logarithmic equation into its equivalent exponential form. The general relationship is: if log_b(y) = x, then b^x = y. Applying this to our original equation, log₅(x²) = 2, we can see that the base b is 5, the exponent x is 2, and the argument y is x². So, converting this to exponential form gives us 5² = x². This is a much simpler equation to solve! It directly relates the base to the argument. This conversion is a fundamental skill when working with logarithms, and it often unlocks the path to a solution. We've transformed a logarithmic equation into a basic algebraic equation that we can solve using familiar methods. So, whether you prefer using the power rule first or converting to exponential form, both paths lead us to a place where we can find the value of x. It's good to know multiple ways to solve a problem, guys, because sometimes one method might click better than another depending on the specific numbers or variables involved.

Solving for x: The Exponential Approach

Okay, let's focus on the exponential form we just derived: 5² = x². This is where the magic really happens, and it’s surprisingly simple. We know that 5² is simply 5 * 5, which equals 25. So, our equation becomes 25 = x². Now, we need to find the value(s) of x that, when squared, give us 25. To do this, we take the square root of both sides of the equation. Remember, when you take the square root of a number, there are always two possible solutions: a positive one and a negative one. So, the square root of x² is x, and the square root of 25 is both +5 and -5. Therefore, we get x = ±5.

This means that there are two possible values for x that satisfy the original logarithmic equation: x = 5 and x = -5. It's super important to remember both the positive and negative roots when solving equations like this. Don't forget that when you square a negative number, you get a positive result. For example, (-5)² = (-5) * (-5) = 25, which is exactly what we need.

Before we declare victory, we must always check our solutions in the original logarithmic equation to ensure they are valid. Why? Because logarithms have domain restrictions. The argument of a logarithm (the part inside the parentheses) must be positive. In our original equation, log₅(x²) = 2, the argument is x². Let's check our solutions:

• For x = 5: The argument is x² = 5² = 25. Since 25 is positive, log₅(25) is defined. And indeed, log₅(25) = 2 because 5² = 25. So, x = 5 is a valid solution.
• For x = -5: The argument is x² = (-5)² = 25. Again, 25 is positive, so log₅(25) is defined. And as we saw, log₅(25) = 2. So, x = -5 is also a valid solution.

Both solutions work perfectly fine! This is a key takeaway: always remember to check for domain restrictions when dealing with logarithms and to consider both positive and negative roots when taking square roots. This careful checking ensures we haven't missed anything or included any invalid answers.

Alternative Path: Using the Power Rule

Now, let's explore the other path we mentioned – using the power rule of logarithms first. Remember, our original equation is log₅(x²) = 2. The power rule states that log_b(M^p) = p * log_b(M). Applying this to our equation, we can bring the exponent 2 down in front of the logarithm:

2 * log₅(x) = 2

This looks a bit different, but it's leading us to the same place. Our goal now is to isolate log₅(x). We can do this by dividing both sides of the equation by 2:

(2 * log₅(x)) / 2 = 2 / 2

Which simplifies to:

log₅(x) = 1

Now, we have a much simpler logarithmic equation. This equation is asking: "To what power must we raise 5 to get x?" And the answer is 1. So, we can convert this into its exponential form. Using the rule if log_b(y) = x, then b^x = y, we have:

• Base b = 5
• Exponent x = 1
• Argument y = x

Plugging these into the exponential form b^x = y, we get:

5¹ = x

And since 5¹ is simply 5, we find that:

x = 5

Wait a minute, guys! This method only gave us one solution, x = 5. What happened to x = -5? This is a super important point and a common pitfall when using the power rule incorrectly. When we apply the power rule log_b(M^p) = p * log_b(M), we are implicitly assuming that M is positive, because the logarithm log_b(M) is only defined for positive values of M. In our original equation, log₅(x²) = 2, the argument is x². As we saw, x² is always positive (or zero) for any real number x. However, when we rewrite it as 2 * log₅(x), we are now restricting the domain. The term log₅(x) requires x to be positive. This means that by applying the power rule in this way, we've inadvertently eliminated the possibility of x being negative from our consideration from that point forward.

To correctly use the power rule and account for both positive and negative values of x, we should use the property log_b(M^p) = p * log_b(|M|) if p is an even integer. In our case, the exponent is 2 (an even integer), so we should rewrite log₅(x²) = 2 as:

2 * log₅(|x|) = 2

Now, dividing by 2, we get:

log₅(|x|) = 1

Converting to exponential form (5¹ = |x|):

|x| = 5

This equation states that the absolute value of x is 5. By definition of absolute value, this means that x can be either 5 or -5. So, x = ±5.

See how using the absolute value correctly with the power rule brings back our two solutions? This highlights the importance of understanding the conditions under which logarithmic properties apply and being mindful of domain restrictions throughout the problem-solving process. It’s a subtle but crucial detail that can make the difference between a correct and an incorrect answer, especially in exam situations where every point counts!

Key Takeaways for Logarithm Problems

Alright team, let's quickly recap the most important lessons learned from tackling this UFRGS 2017 logarithm problem, log₅(x²) = 2. Understanding these points will make you a logarithm ninja, I promise!

First off, always remember the definition of a logarithm. It's the inverse of exponentiation. log_b(y) = x is equivalent to b^x = y. This conversion is your golden ticket to solving many logarithm problems. Don't be afraid to switch between logarithmic and exponential forms; it's a fundamental skill that simplifies things immensely. In our case, converting log₅(x²) = 2 directly to 5² = x² made the problem significantly easier to approach.

Secondly, master the properties of logarithms. The power rule (log_b(M^p) = p * log_b(M)) is incredibly useful for simplifying expressions. However, and this is a big however, be extremely careful when applying it to terms with even exponents like x². Remember that log_b(M) requires M > 0. If your original expression has x² and you rewrite it using the power rule, you must consider the possibility that x could be negative. Using log_b(x²) = 2 * log_b(|x|) is the correct way to handle this and ensures you don't lose potential solutions. This subtlety is often where students make mistakes, so keep it in your mental toolbox!

Third, and this is non-negotiable, always check your solutions against the domain of the original logarithmic equation. The argument of a logarithm must always be positive. In log₅(x²) = 2, the argument is x². This means x² must be greater than 0. Both x = 5 (giving 5² = 25) and x = -5 (giving (-5)² = 25) result in a positive argument, so both are valid. If the original problem had been, say, log₅(x) = 2, then x = 5 would be the only solution because x itself must be positive.

Finally, when you encounter equations that lead to x² = k (where k is a positive constant), remember that there are two possible solutions for x: the positive square root and the negative square root. So, x = ±√k. Don't just default to the positive root; that's how you miss half the answers!

By keeping these points in mind – definition, properties (with caution!), domain checks, and handling square roots – you’ll be well-equipped to tackle a wide variety of logarithm problems, not just on the UFRGS exam but in any math context. Keep practicing, guys, and these concepts will become second nature. You've got this!